Giải bài tập 3.5 trang 32 SBT toán 10 tập 1 kết nối

Trả lời:

a) $sin^{4}\alpha + cos^{4}\alpha = 1 - 2sin^{2}\alpha cos^{2}\alpha$

Xét $sin^{4}\alpha + cos^{4}\alpha$

= $(sin^{2}\alpha)^{2} + (cos^{2}\alpha)^{2}$

= $sin^{2}\alpha + 2sin^{2}\alpha cos^{2}\alpha + cos^{2}\alpha - 2sin^{2}\alpha cos^{2}\alpha$

= $(sin^{2}\alpha + 2sin^{2}\alpha cos^{2}\alpha + cos^{2}\alpha) - 2sin^{2}\alpha cos^{2}\alpha$ 

= $(sin^{2}\alpha + cos^{2}\alpha)^{2} - 2sin^{2}\alpha cos^{2}\alpha$

=  1 - $2sin^{2}\alpha cos^{2}\alpha$

b) $sin^{6}\alpha + cos^{6}\alpha = 1 - 3sin^{2}\alpha cos^{2}\alpha$

Xét $sin^{6}\alpha + cos^{6}\alpha$

= $(sin^{2}\alpha)^{3} + (cos^{2}\alpha)^{3}$ 

= $(sin^{2}\alpha + cos^{2}\alpha).(sin^{4}\alpha - sin^{2}\alpha cos^{2}+cos^{4}\alpha)$

= $(sin^{2}\alpha + cos^{2}\alpha).(sin^{4}\alpha + 2sin^{2}\alpha cos^{2}\alpha + cos^{4}\alpha - 3sin^{2}\alpha cos^{2}\alpha)$

= 1.$(sin^{4}\alpha + 2sin^{2}\alpha cos^{2}\alpha + cos^{4}\alpha - 3sin^{2}\alpha cos^{2}\alpha)$

= $(sin^{2}\alpha + cos^{2}\alpha)^{2} - 3sin^{2}\alpha cos^{2}\alpha$

= 1 - $3sin^{2}\alpha cos^{2}\alpha$

c*) $\sqrt{sin^{4}\alpha + 6cos^{2}\alpha + 3} + \sqrt{cos^{4}\alpha + 4sin^{2}\alpha}$ = 4

Xét $\sqrt{sin^{4}\alpha + 6cos^{2}\alpha + 3} + \sqrt{cos^{4}\alpha + 4sin^{2}\alpha}$ 

= $\sqrt{(1-cos^{2}\alpha)^{2} + 6cos^{2}\alpha + 3} + \sqrt{(cos^{2})^{2} + 4sin^{2}\alpha}$

= $\sqrt{4 + 4cos^{2}\alpha + cos^{4}\alpha} + \sqrt{1 + 2sin^{2}\alpha + sin^{4}\alpha}$

= $(2 + cos^{2}\alpha) + (1 + sin^{2}\alpha)$

= 4